Question: $f(x, y, z) = x^2z^4 - z^2y^3 + x$ What is the Laplacian of $f(x, y, z)$ ? $\nabla^2 f = $
Explanation: The Laplacian of a scalar field $f$ is the sum of each of its second partial derivatives. $\nabla^2 f = \dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2} + \dfrac{\partial^2 f}{\partial z^2}$ [What does it mean to square the gradient?] Let's find the second partial derivatives of $f$ ! $\begin{aligned} f_{xx} &= \dfrac{\partial}{\partial x} \left[ \dfrac{\partial f}{\partial x} \right] \\ \\ &= \dfrac{\partial}{\partial x} \left[ 2xz^4 + 1 \right] \\ \\ &= 2z^4 \\ \\ f_{yy} &= \dfrac{\partial}{\partial y} \left[ \dfrac{\partial f}{\partial y} \right] \\ \\ &= \dfrac{\partial}{\partial y} \left[ -3z^2y^2 \right] \\ \\ &= -6z^2y \\ \\ f_{zz} &= \dfrac{\partial}{\partial z} \left[ \dfrac{\partial f}{\partial z} \right] \\ \\ &= \dfrac{\partial}{\partial z} \left[ 4x^2z^3 - 2zy^3 \right] \\ \\ &= 12x^2z^2 - 2y^3 \end{aligned}$ The Laplacian is $\nabla^2 f = f_{xx} + f_{yy} + f_{zz}$. Therefore: $\nabla^2 f(x, y, z) = 2z^4 - 6z^2y + 12x^2z^2 - 2y^3$